Physics 205B Quiz 5, spring semester 2013
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 18.11(b), Comprehensive Problem 18.99
A 2012 Hyundai Sonata uses 470 mA of current from its 12 V battery while parked and with the engine turned off[*]. Over a 24 hour period, the battery will have supplied __________ of energy.
(A) 6.5×10-5 J.
(B) 3.4×103 J.
(C) 4.9×105 J.
(D) 2.2×106 J.
[*] Phil B., "Car Battery Goes Dead after A Few Days," http://www.instructables.com/id/Car-Battery-Goes-Dead-after-A-Few-Days/.
Correct answer: (C)
Although there are number of equations that could be used to calculate the correct answer, here a purely unit analysis method is used. Given:
470 mA = 470×10-3 C/s;
12 V = 12 J/C;
24 h = 86,400 s.
Seeking an answer in joules, the trial solution is:
? J = (470×10-3 C/s)a·(12 J/C)b·(86,400 s)c.
From inspection, the exponents a, b, and c are all unity, such that the units of C and s cancel out:
? J = (470×10-3 C/s)·(12 J/C)·(86,400 s) = 487,296 J = 4.9×105 J.
Response (A) is (470×10-3 A)·(12 V)/(86,400 s); response (B) is (470×10-3 A)·(86,400 s)/(12 V); and response (C) is (12 V)·(86,400 s)/(470×10-3 A).
Section 30882
Exam code: quiz05c4Rb
(A) : 6 students
(B) : 8 students
(C) : 14 students
(D) : 4 students
Success level: 44%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.45
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